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-Dominant- [34]
4 years ago
9

I need answer ASAP!!!!

Mathematics
1 answer:
frutty [35]4 years ago
7 0
The answer is 1 since both expressions are equal.
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Three regular hexagons are put together to make a larger shape. The perimeter of one of the regular hexagons is 18cm. Find the p
pogonyaev

Answer:

steps below

Step-by-step explanation:

2 kinds of larger shapes can be make

kind 1: 18 x 4 x 3 = 216 cm

kind 2: 18 x (5+5+4) = 252 cm

5 0
3 years ago
If f(x) = <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%283%2Bx%29%7D%7B%28x-3%29%7D" id="TexFormula1" title="\frac{(3+x)}{(x-3)}
kati45 [8]

Answer:

f(a) = \frac{(a+5)}{(a-1)}

Step-by-step explanation:

Given : f(x) = \frac{(3+x)}{(x-3)}

In order to find f(a+2) we will substitute (a+2) for x in the given function, that is :

f(a) = \frac{(3+a+2)}{(a+2-3)}

f(a) = \frac{(a+5)}{(a-1)}

4 0
4 years ago
Orange M&amp;M’s: The M&amp;M’s web site says that 20% of milk chocolate M&amp;M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0
4 years ago
-3(-4b - 7n) use n = 8 and b = 2 show works plsssssss
Lilit [14]

Answer:

192

Step-by-step explanation:

-3(-4(2) - 7(8))

= -3(-8 - 56)

= -3(-64)

= 192

*Negative x negative = positive

7 0
3 years ago
Jim provides photos for two online sites: site A and site B. Site A pays $0.85 for every photo Jim provides. The amount in dolla
yarga [219]
We can use the same equation to figure out with only substituting the amount he was paid per photo.

Site A    y = $0.85x      =  $0.85(5)  =  $4.25
Site B    y = $0.40x      =  $0.40(5)  =  $2.00

We can see that Jim was paid $2.25 more at site A than Site B.
6 0
3 years ago
Read 2 more answers
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